Question: What is the value of the following logarithm? $\log_{32} 2$
Answer: If $b^y = x$ , then $\log_{b} x = y$ Notice that $2$ is the fifth root of $32$ That is, $\sqrt[5]{32} = 32^{1/5} = 2$ Thus, $\log_{32} 2 = \dfrac{1}{5}$.